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Factoring Polynomials
Sometimes when we try to solve or simplify an equation or expression involving polynomials the way that
it looks can hinder our progress in finding a solution. Factorization is a means of rewriting an expression in a
way that can make it much easier to work with and simplify.
1
GCF
We begin with the easiest and often the most overlooked step in factoring: finding a greatest common factor
(gcf). Take for example the following binomial:
6x + 26
Notice that the coefficient on x and the constant term are both multiples of 2. We can use this fact to simplify
the expression
6x + 26 = (2 · 3)x + (2 · 13)
= 2 · (3x) + 2(13)
(1.1)
= 2 · (3x + 13)
Now sometimes the common factor includes variables, for example
121x2 y + 33y 2 = 11y(11x2 + 3y)
(1.2)
Looking for a GCF is always the first thing one should do when factoring.
2
2-Term
The 2-term formulae are very simple to memorize, the difficulty is that they can be buried in the expression.
For this reason we will do little more than state them.
2.1
Difference of Squares
The difference of squares is an easy pattern to memorize:
x2 − y 2 = (x − y)(x + y)
(2.1)
A very simple example may look much like this:
2.2
2.3
2.4
x2 − 4 = (x − 2)(x + 2)
(2.2)
x3 − y 3 = (x − y)(x2 + xy + y 2 )
(2.3)
x3 + y 3 = (x + y)(x2 − xy + y 2 )
(2.4)
Difference of Cubes
Sum of Cubes
Conclusion
As easy as these patterns are, sometimes it can be hidden in much stranger looking expressions like the following
which can require quite a bit of algebraic manipulation.
√
√
√
4
5
8x 2 − 18 xy 6 = 2 · 4 x · x 2 − 2 · 9 x · y 6
√
= 2 x(4x2 − 9(y 3 )2 )
(2.5)
√
= 2 x((2x)2 − (3y 3 )2 )
√
= 2 x(2x − 3y 3 )(2x + 3y 3 )
1
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3-term
3.1
1-Leading Coefficient
Now we will turn our attention to factoring trinomials of the form
x2 + bx + c
(3.1)
To do this we simply
1. Find two numbers p and q such that p + q = b and pq = c
2. Write x2 + bx + c = (x + p)(x + q)
This could be rather complicated but since we are usually talking about the case where b and c are integers,
and we are usually only willing to factor if we can find integers p and q, it really isn’t too bad.
In such a case to reduce the number of “guesses” required to find such a p and q, we start by determining the
pairs of factors of c and then from there we check the sums to find which pair (if any) sums to b. For example:
x2 − x − 6
(3.2)
Now we look at the pairs of factors for -6:
(−1, 6), (1, −6), (−2, 3), (2, −3)
Next do the sums
5, −5, 1, −1
Since -1 = b we determine that the desired (p, q) is (2,-3) and hence
x2 − x − 6 = (x − 3)(x + 2)
3.2
(3.3)
Non-1 Leading Coefficient
Naturally the next step is to generalize the above procedure to trinomials of the form
ax2 + bx + c where a 6= 0
(3.4)
(If a = 1 then we just have the case covered in section 3.1)
The first question we should ask ourselves is can we factor out a? Remember always look for a GCF before
doing anything else!
3.2.1
Factoring out the Offending Coefficient
In the case where a is a factor of both b and c, that is where b = aj and c = ak for some integers j and k then
we can reduce it to the previous case by way of
ax2 + bx + c = ax2 + (aj)x + (ak) = a(x2 + jx + k)
(3.5)
This might look a little strange, so a concrete, simple example is helpful:
5x2 + 25x + 30 = 5(x2 + 5x + 6)
= 5(x + 3)(x + 2)
2
(3.6)
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(Fairly) General Trinomial Factoring
If we can’t simply factor out the leading coefficient, we must work on a slightly more complicated task to factor
ax2 + bx + c:
Let’s factor the following trinomial:
6x2 + 29x + 35
The first step is to find a pair of numbers (p, q) such that:
pq = ac
p+q =b
So in our example we need to find two numbers whose product is 6 · 35 = 210 and whose sum is 29.
Let’s try a few pairs of factors for 210:
6 · 35 = 210
21 · 10 = 210
14 · 15 = 210
but 6 + 35 = 41 6= 29
but 21 + 10 = 31 6= 29
and 14 + 15 = 29
So that pair is no good.
So that pair is not good either.
So (14,15) is our pair.
Now we use our two numbers to decompose the middle term, group and factor. This sounds a little vague,
but working this example will make things clear.
6x2 + 29x + 35 = 6x2 + (14x + 15x) + 35
Decompose the middle term
2
= (6x + 14x) + (15x + 35)
Regroup terms
= (2x)(3x + 7) + 5(3x + 7)
Factor the groups
= (3x + 7)(2x + 5)
Factor out the common binomial factor
∴ 6x2 + 29x + 35 = (3x + 7)(2x + 5)
With this procedure the hardest part is finding our pair of numbers. Luckily finding such numbers gets
quicker with practice.
3
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3.3
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Perfect Square Trinomial
Given a trinomial of the form
a2 x2 + bx + c2 where b = 2ac
(3.7)
then
a2 x2 + bx + c2 = a2 x2 + 2acx + c2
(3.8)
= (ax + c)2
Example:
4x2 + 12x + 9 = 4x2 + 2 · 6x + 9
= 22 x2 + 2 · 2 · 3x + 32
= (2x + 3)2
If you do not see that a trinomial is a perfect square, that is okay. Just factor it using one of the factoring
procedures and you will sucessfully factor the trinomial. It is desirable to spot patterns such as the perfect
square because it can greatly speed up factoring.
3.4
Quadratic Formula
We should be familiar with the quadratic formula which states if ax2 + bx + c = 0 then x =
let α and β be the two not necessarily distinct solutions to ax2 + bx + c = 0, then
ax2 + bx + c = a(x − α)(x − β)
√
−b± b2 −4ac
.
2a
We
(3.9)
The resulting factors are not necessarily in the most aestetically pleasing form but they are fast to find and
it may be easier to polish it afterward than it is to do it the long way. Let’s try an example:
−14x2 + 31x − 15
Now we will use the quadratic formula:
−31 +
(α, β) =
p
312 − 4(−14)(−15) −31 −
,
2(−14)
!
p
312 − 4(−14)(−15)
2(−14)
...
!
√
√
−31 + 121 −31 − 121
,
−28
−28
=
(3.10)
...
=
5 3
,
7 2
So
5
3
−14x2 + 31x − 15 = −14 x −
x−
7
2
5
3
= −7 x −
2 x−
7
2
(3.11)
= (−7x + 5)(2x − 3)
In this case it was probably easier to factor the trinomial using our original method, but sometimes it can be
difficult to find our pair (p,q). If finding p and q is difficult, then the quadratic formula gives us an alternative
method.
4
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4-term
Sometimes it is best to group terms in some slightly more creative ways. These are just a few patterns that
may be useful, many more can be found by combining the methods that we have discussed already
4.1
3 & 1 Term Grouping
3 & 1 Term Grouping is often used when we have problems with two variables. We’ll start with an example:
x2 + 6x + 9 − y 2 = (x2 + 6x + 9) − y 2
= (x + 3)2 − y 2
(4.1)
= ((x + 3) + y)((x + 3) − y)
= (x + y + 3)(x − y + 3)
In such that example we found a perfect square trinomial inside of the tetranomial, factored it, then noticed
that we had a difference of squares and then factored that. Let’s look at even a stranger example:
x2 + 10x − 4y 2 + 12y + 14 = x2 + 10x − 4y 2 + 12y + 25 − 9
= x2 + 10x + 25 − 4y 2 + 12y − 9
= (x2 + 10x + 25) − (4y 2 − 12y + 9)
= (x + 5)2 − (2y − 3)2
(4.2)
= ((x + 5) + (2y − 3))((x + 5) − (2y − 3))
= (x + 2y + 2)(x − 2y + 8)
We can generalize this pattern as follows:
a2 x2 + 2abx − α2 y 2 − 2αβy + (b2 − β 2 ) = (ax + αy + b + β)(ax − αy + b − β)
(4.3)
Given the complexity of these pattern, it is often better just to remember the reasoning rather than the
generalized formula.
4.2
2 & 2 Term Grouping
2 & 2 Term Grouping is typically used with expressions in 1 variable with a degree greater than 2. We shall
leave the explanation to an example.
x5 + x4 + x + 1 = (x5 + x4 ) + (x + 1)
= x4 (x + 1) + 1(x + 1)
(4.4)
4
= (x + 1)(x + 1)
5
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